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Multiple modulus inequalities

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Multiple modulus inequalities  [#permalink]

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New post 19 Dec 2009, 02:01
1
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Hi guys

Need to know how to solve multiple modulus inequalities

Ex - Solve |x-10| - |2x+3| <= |5x-10|

I do understand that we need to first find the roots, which happen to be 10, -3/2 and 2 in this case

we can then divide them into intervals -

x<-3/2
-3/2<x<2
2<x<10
x>10

what do we do post this division into intervals?
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Re: Multiple modulus inequalities  [#permalink]

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New post 19 Dec 2009, 03:47
3
zaarathelab wrote:
Hi guys

Need to know how to solve multiple modulus inequalities

Ex - Solve |x-10| - |2x+3| <= |5x-10|

I do understand that we need to first find the roots, which happen to be 10, -3/2 and 2 in this case

we can then divide them into intervals -

x<-3/2
-3/2<x<2
2<x<10
x>10

what do we do post this division into intervals?


If we dot he way you are proposing:

Yes, there are 3 check point at which absolute values in the example change their sign: -3/2, 2, and 10 (te values of x for which the expression in || equals to zero). Then you should expand the modulus in these ranges:

A. \(x<-\frac{3}{2}\) --> \(|x-10|-|2x+3|\leq{|5x-10|}\) --> \(-(x-10)+(2x+3)\leq{-(5x-10)}\) --> \(6x\leq{-3}\) --> \(x\leq{-\frac{1}{2}\) --> as we considering the range when \(x<-\frac{3}{2}\), then finally we should write \(x<-\frac{3}{2}\):
----(\(-\frac{3}{2}\))---------------------------------

B. \(-\frac{3}{2}\leq{x}\leq{2}\) --> \(-(x-10)-(2x+3)\leq{-(5x10)}\) --> \(2x<=3\) --> \(x<={\frac{3}{2}\) --> \({-\frac{3}{2}}\leq{x}\leq{\frac{3}{2}}\);
----(\(-\frac{3}{2}\))----(\(\frac{3}{2}\))-------------------------

C. \(2<x<10\) --> \(-(x-10)-(2x+3)\leq{5x-10}\) --> \(17\leq{8x}\) --> \(x\geq{\frac{17}{8}}\) --> \({\frac{17}{8}}\leq{x}<10\);
---------------------(\(\frac{17}{8}\))--------(\(10\))----

D. \(x\geq{10}\) --> \((x-10)-(2x+3)\leq{5x-10}\) --> \(-3\leq{6x\) --> \(x\geq{-\frac{1}{2}}\) --> \(x\geq{10}\);
---------------------(\(\frac{17}{8}\))--------(\(10\))----

After combining the ranges we've got, we can conclude that the ranges in which the given inequality holds true are:
\(x\leq{\frac{3}{2}}\) and \({\frac{17}{8}}\leq{x}\)
----(\(-\frac{3}{2}\))----(\(\frac{3}{2}\))----(\(\frac{17}{8}\))--------(\(10\))-----


One important detail when you define the critical points (10, -3/2 and 2) you should include them in either of intervals you check. So when you wrote:
x<-3/2
-3/2<x<2
2<x<10
x>10

You should put equal sign for each of the critical point in either interval. For example:
x=<-3/2
-3/2<x<=2
2<x<1=0
x>10

Notice, that in solution I put the equal sign in different way, but it doesn't matter. It's important just to consider the cases when x equals to the critical points and include them when expanding given inequality.
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Re: Multiple modulus inequalities  [#permalink]

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New post 01 Feb 2012, 02:31
Could somebody explain this in detail, please?

I didn't get the concept of roots here, it doesn't make the equation zero - isn't it?
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Re: Multiple modulus inequalities  [#permalink]

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New post 01 Feb 2012, 02:47
bsaikrishna wrote:
Could somebody explain this in detail, please?

I didn't get the concept of roots here, it doesn't make the equation zero - isn't it?


When x<=0 then |x|=x and if x>0 then |x|=-x.

For |x-10|: when x<=10 then |x-10|=-(x-10) (because if x<=10 then x-10<=0) and when x>10 then |x-10|=x-10 (because if x>10 then x-10>0). So, zaarathelab calls "roots" (you can also call "critical point" or "check point") the values of x for which the expression in || equals to zero: in two different cases when x<that value and x>that value the absolute values expands with different sign (it switches the sign at this value).

Check Absolute Values chapter of Math Book for more: math-absolute-value-modulus-86462.html

Hope it helps.
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Re: Multiple modulus inequalities  [#permalink]

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New post Updated on: 01 Feb 2012, 03:05
Hi,

I solved the modulus by taking the positive & negative scenarios of each modulus. I solved them separately & arrived at 1/2 <= x & x<= -1/2.

Am I on the right track or messed up some where??
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Originally posted by boomtangboy on 01 Feb 2012, 02:55.
Last edited by boomtangboy on 01 Feb 2012, 03:05, edited 1 time in total.
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Re: Multiple modulus inequalities  [#permalink]

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New post 01 Feb 2012, 02:59
boomtangboy wrote:
Hi,

I solved the modulus by taking the positive & negative scenarios of each modulus. I solved them separately & arrived at 1/2 <= x & x<= -1/2.

Am I on the right track or messed up some where??


The answer to the initial question is x<=3/2 or 17/8<=x, so "1/2 <= x & x<= -1/2" is not right. Check my previous post: you should check different cases for different ranges to get the correct answer.
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Collection of Questions:
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Re: Multiple modulus inequalities  [#permalink]

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New post 04 Oct 2012, 11:45
Bunuel wrote:
zaarathelab wrote:
Hi guys

Need to know how to solve multiple modulus inequalities

Ex - Solve |x-10| - |2x+3| <= |5x-10|

I do understand that we need to first find the roots, which happen to be 10, -3/2 and 2 in this case

we can then divide them into intervals -

x<-3/2
-3/2<x<2
2<x<10
x>10

what do we do post this division into intervals?


If we dot he way you are proposing:

Yes, there are 3 check point at which absolute values in the example change their sign: -3/2, 2, and 10 (te values of x for which the expression in || equals to zero). Then you should expand the modulus in these ranges:

A. \(x<-\frac{3}{2}\) --> \(|x-10|-|2x+3|\leq{|5x-10|}\) --> \(-(x-10)+(2x+3)\leq{-(5x-10)}\) --> \(6x\leq{-3}\) --> \(x\leq{-\frac{1}{2}\) --> as we considering the range when \(x<-\frac{3}{2}\), then finally we should write \(x<-\frac{3}{2}\):
----(\(-\frac{3}{2}\))---------------------------------

B. \(-\frac{3}{2}\leq{x}\leq{2}\) --> \(-(x-10)-(2x+3)\leq{-(5x10)}\) --> \(2x<=3\) --> \(x<={\frac{3}{2}\) --> \({-\frac{3}{2}}\leq{x}\leq{\frac{3}{2}}\);
----(\(-\frac{3}{2}\))----(\(\frac{3}{2}\))-------------------------

C. \(2<x<10\) --> \(-(x-10)-(2x+3)\leq{5x-10}\) --> \(17\leq{8x}\) --> \(x\geq{\frac{17}{8}}\) --> \({\frac{17}{8}}\leq{x}<10\);
---------------------(\(\frac{17}{8}\))--------(\(10\))----

D. \(x\geq{10}\) --> \((x-10)-(2x+3)\leq{5x-10}\) --> \(-3\leq{6x\) --> \(x\geq{-\frac{1}{2}}\) --> \(x\geq{10}\);
---------------------(\(\frac{17}{8}\))--------(\(10\))----

After combining the ranges we've got, we can conclude that the ranges in which the given inequality holds true are:
\(x\leq{\frac{3}{2}}\) and \({\frac{17}{8}}\leq{x}\)
----(\(-\frac{3}{2}\))----(\(\frac{3}{2}\))----(\(\frac{17}{8}\))--------(\(10\))-----


One important detail when you define the critical points (10, -3/2 and 2) you should include them in either of intervals you check. So when you wrote:
x<-3/2
-3/2<x<2
2<x<10
x>10

You should put equal sign for each of the critical point in either interval. For example:
x=<-3/2
-3/2<x<=2
2<x<1=0
x>10

Notice, that in solution I put the equal sign in different way, but it doesn't matter. It's important just to consider the cases when x equals to the critical points and include them when expanding given inequality.



Hi Brunel,

How do we decide the signs in front of the modulus in the equation. :(
Eg:: |x-10| - |2x+3| <= |5x-10|......say we are looking in an interval - 3/2 to 2. then based on what concentration do we put +or - sign in front of the modulus.


and please explain >> step 1----------x = sqrt (x^2) ?
< step w--------------- => x = |x| ? How do this. please explain concept behind this math.

Thanks!!
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Re: Multiple modulus inequalities  [#permalink]

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New post 05 Oct 2012, 04:50
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Maverick04308 wrote:
Hi Brunel,

How do we decide the signs in front of the modulus in the equation. :(
Eg:: |x-10| - |2x+3| <= |5x-10|......say we are looking in an interval - 3/2 to 2. then based on what concentration do we put +or - sign in front of the modulus.


and please explain >> step 1----------x = sqrt (x^2) ?
< step w--------------- => x = |x| ? How do this. please explain concept behind this math.

Thanks!!


Explained here: multiple-modulus-inequalities-88174.html#p1037473

Hope it helps.
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: Multiple modulus inequalities  [#permalink]

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